Problem 45321. Kolakoski Sequence
Solution Stats
Problem Comments
-
3 Comments
So it's kind of like Kolakoski; the number still gives the length at pos n, like A000002. However, the following number is decided according to a base named "a". So, if a=[4,2,3], the remainder 0,1 or 2 (since a has length 3) will be mapped to 4,2,3 in this order.
I recommend solving the problem in the description if you haven't already. You can reuse your code here. You will need to make some changes, but it will be easier than solving this one from zero. Numberphile has a video about it if you are still stuck.
I fail to understand some of the examples, where the initial terms of the output are not the same as the input (a) -
a = [1,3,1,2];
x=30;
y=[1, 3, 3, 3,...] %first 4 terms of y not equal to a
a = [2,1];
x=10;
y=[2, 2, 1, ...] %first 2 terms of y not equal to a
Solution Comments
Show commentsProblem Recent Solvers12
Suggested Problems
-
Basic electricity in a dry situation
544 Solvers
-
Convert a vector into a number
585 Solvers
-
239 Solvers
-
Pandigital number n°1 (Inspired by Project Euler 32)
80 Solvers
-
313 Solvers
More from this Author165
Problem Tags
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!